Posted by lulu78 on Friday, October 15, 2010 at 5:33am.
0.1m aqueous solution of HCl and 0.1m aqueous solution of C6H12O6.
calculate the boiling point of each
i know the eq. dTb=Kbm
but i have no temperatures. this is the question
- CHEMISTRY- boiling point - drwls, Friday, October 15, 2010 at 8:17am
It is an aqueous solution. You know that the boiling point with no HCl is 100 C. Use Kb and the molarity to compute the change in B.P.
- CHEMISTRY- boiling point - lulu78, Friday, October 15, 2010 at 10:08pm
can you show me. i'm confused.
- CHEMISTRY- boiling point - DrBob222, Friday, October 15, 2010 at 11:36pm
delta T = i*Kb*molality
i is the van't Hoff factor, which for HCl is 2 (it dissolves into two particles).
delta T = 2*0.512*0.1 = 0.102
Therefore, the new boiling point for water must be 100 C + 0.102 = ??
For the sugar solution,
i = 1 since it doesn't ionize.
delta T = i*Kb*m
delta T = 1*0.512*0.1 = 0.0512
New boiling point will be
100 C + 0.0512.
The problem is hoping you will see the difference between the two; i.e., the HCl ionizes to give two particles and has just twice the effect as the other one although the molality is the same.
- CHEMISTRY- boiling point - lulu78, Saturday, October 16, 2010 at 2:13am
is the Kb 0.51 just standard BP of water? i think thats where i'm getting lost.
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